10 ml of gaseous hydrocarbon on combution gives 20ml of `CO_(2)`
and 30ml of `H_(2)O(g)`. The hydrocarbon is :-

Options

A. `C_(4)H_(5)`
B. `C_(2)H_(6)`
C. `C_(4)H_(8)`
D. `C_(4)H_(10)`

Explanation

To solve the problem, we need to determine the formula of the hydrocarbon based on the volumes of carbon dioxide (CO₂) and water vapor (H₂O) produced during its combustion. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the combustion reaction
Assume the hydrocarbon is represented as \( C_xH_y \). The general combustion reaction can be written as:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Balance the combustion reaction
From the combustion of \( C_xH_y \), we can deduce the following:
- For every molecule of \( C_xH_y \), \( x \) molecules of \( CO_2 \) and \( \frac{y}{2} \) molecules of \( H_2O \) are produced.
- The balanced equation can be expressed as:
\[ C_xH_y + O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O \]

### Step 3: Relate the volumes of gases
According to the problem:
- 10 ml of the hydrocarbon produces 20 ml of \( CO_2 \) and 30 ml of \( H_2O \).
- From the volumes, we can establish the following relationships:
- The volume of \( CO_2 \) produced from 10 ml of hydrocarbon is \( 10x \) ml.
- The volume of \( H_2O \) produced from 10 ml of hydrocarbon is \( 5y \) ml (since \( \frac{y}{2} \) is produced per 1 ml of hydrocarbon).

### Step 4: Set up equations based on the given volumes
From the information provided:
1. \( 10x = 20 \) (for \( CO_2 \))
2. \( 5y = 30 \) (for \( H_2O \))

### Step 5: Solve for \( x \) and \( y \)
1. From the first equation:
\[ x = \frac{20}{10} = 2 \]
2. From the second equation:
\[ y = \frac{30}{5} = 6 \]

### Step 6: Write the molecular formula of the hydrocarbon
Now that we have determined \( x \) and \( y \):
- The hydrocarbon is \( C_2H_6 \).

### Conclusion
The hydrocarbon that produces 20 ml of \( CO_2 \) and 30 ml of \( H_2O \) upon combustion is \( C_2H_6 \) (ethane).

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